Binary Search Tree

What is a Binary Tree?
 Property 1: Each node can have up to two
successor nodes.

What is a Binary Tree?
• Property 2: a unique path exists from the root
to every other node
Not a valid binary tree!

Binary Tree
typedef struct tnode *ptnode;
typedef struct node {
short int key;
ptnode right, left;
} ;

Basic Terminology
 The successor nodes of a node are called its children
 The predecessor node of a node is called its parent
 The "beginning" node is called the root (has no parent)
 A node without children is called a leaf

Basic Terminology
 Nodes are organize in levels (indexed from 0).
 Level (or depth) of a node: number of edges in the path
from the root to that node.
 Height of a tree h: #levels = L
(some books define h as
number of levels-1).
 Full tree: every node has exactly
two children and all the leaves are
on the same level.
not full!

What is the max number of nodes
at any level l ?
The maximum number of nodes at any level l is less
than or equal to 2l where l=0, 1, 2, 3,….. L-1
where l=0,1,2, ...,L-1
£ 20
£ 21
£ 22
£ 23

What is the total number of nodes N
of a full tree with height h?
h -
h
0 1 1
= + + + = -
2 2 ... 2 2 1
Derived according to the geometric series:
-
1
+ + + = å =
- -
0 1 1 1
1
... n
0
n
n i x
x
i
N
x x x x
-
=
l=0 l=1 l=h-1

What is the height h
of a full tree with N nodes?
2 - 1
=
Þ 2 = +
1
Þ = + ®
log( 1) (log )
h
h
N
N
h N O N

Why is h important?
 The Tree operations like insert, delete, retrieve etc.
are typically expressed in terms of the height of the
tree h.
 So, it can be stated that the tree height h determines
running time!

 What is the mmaaxx hheeiigghhtt of a tree
with N nodes?
N (same as a linked list)
 What is the mmiinn hheeiigghhtt of a tree with
N nodes?
log(N+1)

Binary Search
 Suppose DATA is an array which is sorted in increasing
numerical order and we want to find the location LOC of a given
ITEM in DATA. Then there is an extremely efficient searching
algorithm called Binary Search.

Binary Search
 Consider that you want to find the location of some word in a
dictionary.
 I guess you are not fool enough to search it in a linear way or in
other words apply linear search. That is no one search page by
page from the start to end of the book.
 Rather, I guess you will open or divide the dictionary in the
middle to determine which half contains the word.
 Then consider new probable half and open or divide that half in
the middle to determine which half contains the word. This
process goes on.
 Eventually you will find the location of the word and thus you
are reducing the number of possible locations for the word in the
dictionary.

How to search a binary tree?
1. Start at the root
2. Search the tree level by level,
until you find the element you
are searching for or you reach
a leaf.
Is this better than searching a linked list?
No  O(N)

Binary Search Trees
 Binary Search Tree Property:
The value stored at
a node is greater than
the value stored at its
left child and less than
the value stored at its
right child

Binary Search Trees
 In a BST, the value
stored at the root of
a subtree is greater
than any value in its
left subtree and less
than any value in its
right subtree!

Binary Search Trees
Where is the smallest
element?
Ans: leftmost element
Where is the largest
element?
Ans: rightmost element

How to search a binary search
tree?
1. Start at the root
2. Compare the value of
the item you are
searching for with the
value stored at the root
3. If the values are equal,
then item found;
otherwise, if it is a leaf
node, then not found

tree?
4. If it is less than the value
stored at the root, then
search the left subtree
5. If it is greater than the
value stored at the root,
then search the right
subtree
6. Repeat steps 2-6 for the
root of the subtree
chosen in the previous
step 4 or 5

tree?
Is this better than searching
a linked list?
Yes !!   O(logN)

Difference between BT and BST
 A binary tree is simply a tree in which each node can have
at most two children.
 A binary search tree is a binary tree in which the nodes are
assigned values, with the following restrictions :
1. No duplicate values.
2. The left subtree of a node can only have values less than
the node
3. The right subtree of a node can only have values greater
than the node and recursively defined
4. The left subtree of a node is a binary search tree.
5. The right subtree of a node is a binary search tree.

Binary Tree Search Algorithm
 Let x be a node in a binary search tree and k is the
value, we are supposed to search.
 Then according to the binary search tree property we
know that: if y is a node in the left subtree of x, then
y.key <= x.key. If y is a node in the right subtree of x,
then y.key>=x.key. (x.key denotes the value at node x)
 To search the location of given data k, the binary search
tree algorithm begins its search at the root and traces
the path downward in the tree.
 For each node x it compares the value k with x.key. If
the values are equal then the search terminates and x is
the desired node.

 If k is smaller than x.key, then the search continues in
the left subtree of x, since the binary search tree
property implies that k could not be in the right subtree.
 Symmetrically, if k is larger than x.key, then the search
continues in the right subtree.
 The nodes encountered during the recursion form a
simple path downward from the root of the tree, and
thus the running time of TREE-SEARCH is O(h), where h
is the height of the tree.

TREE-SEARCH(x,k)
1.If x==NIL or k==x.key
2. return x
3.If k < x.key
4. return TREE-SEARCH(x.left,k)
5.else return TREE-SEARCH(x.right,k)

6 6
Example: key =4 then find(4)
2 2 9 9
1 1 4 4 8 8
Call TREE-SEARCH(x,k)

6 6
Example: k =4 and x=root
2 2 9 9
1 1 4 4 8 8
Call TREE-SEARCH(root,4)
Now x.key=6 then k<x.key
So call TREE-SEARCH(x.left,k)
x=root

6 6
Example: k =4 and x=root.left
2 2 9 9
1 1 4 4 8 8
Call TREE-SEARCH(x,4)
Now x.key=2 then k>x.key
TREE-SEARCH(x.right,k)
x=root.left

6 6
Example: k =4 and x=root.left.right
2 2 9 9
1 1 4 4 8 8
Call TREE-SEARCH(x,4)
Now x.key=4 then k=x.key
Search terminates
x=root.left.right

6 6
Example: k =4 and x=root.left.right
2 2 9 9
1 1 4 4 8 8
Now x.key=4 then k=x.key
Search terminates and
x=root.left.right is the desired
node or location
x=root.left.right

Animation of How Works in a
Sorted Data Array

BST - Pseudo code
if the tree is empty
return NULL
else if the key value in the node(root) equals the target
return the node value
else if the key value in the node is greater than the target
return the result of searching the left subtree
else if the key value in the node is smaller than the target
return the result of searching the right subtree

Search in a BST: C code
Ptnode search(ptnode root,
int key)
{
/* return a pointer to the node that
contains key. If there is no such
node, return NULL */
if (!root) return NULL;
if (key == root->key) return root;
if (key < root->key)
return search(root->left,key);
return search(root->right,key);
}

Minimum Key or Element
 We can always find an element in a binary search tree
whose key is minimum by following the left children from
the root until we encounter a NIL.
 Otherwise if a node x has no left subtree then the value
x.key contained in root x is the minimum key or element.
The procedure for finding the minimum key:
 TREE-MINIMUM(x)
1. while x.left ≠ NIL
2. x = x.left
3. return x

Maximum Key or Element
 We can always find an element in a binary search tree
whose key is maximum by following the right children from
the root until we encounter a NIL.
 Otherwise if a node x has no right subtree then the value
x.key contained in root x is the maximum key or element.
The procedure for finding the maximum key:
 TREE-MAXIMUM(x)
1. while x.right ≠ NIL
2. x = x.right
3. return x

Insert a value into the BST
 To insert a new value v into a binary search tree T, we use
the procedure TREE-INSERT.
 The procedure takes a node z for which z.key=v , z.left=NIL
and z.right=NIL.
 It modifies T and some of the attributes of z in such a way
that it inserts z into an appropriate position in the tree

 Suppose v is the value we want to insert and z is the node (New
or NIL) we are supposed to find to insert the value v.
z.p denotes the parent of z.
 X is a pointer that traces a simple path downward the tree and y
is the trailing pointer as the parent of x. T.root denote the root
of the tree.
 Now the intention is to find a new or NIL node that will satisfy
the BST property after placing the value v. The procedure first
consider x as the root of the tree thus the parent of the root
y=NIL.
 In steps 3 to 7 the procedure causes the two pointer y and x to
move down the tree, going left or right depending on the
comparison of z.key with x.key, until x becomes NIL.

 Now this NIL occupies the position z, where we wish to place the
input item.
 At this time we need y the parent of the desired node. This is why
at step four we always storing the parent of current node x while
moving downward. At the end of step 7 (in step 8) we make this
node the parent of z (z.p).
 From steps 9 to 11:
 Now if tree is empty (y==NIL) then create a root node with the
new key(T.root=z)
 If the value v is less than the value of the parent(z.key<y.key)
then make it as the left-child of the parent(y.left=z)
 If the value v is greater than the value of the parent(z.key>y.key)
then make it as the right-child of the parent(y.right=z)

 TREE-INSERT(T , z)
1. y=NIL
2. x= T.root
3. While x ≠ NIL
4. y=x
5. if z.key<x.key
6. x=x.left
7. else x=x.right
8. z.p=y
9. if y== NIL
10. T.root = z
11.elseif z.key < y.key
12. y.left=z
13.else y.right = z

BST Insertion Algorithm
6 6
2 2 9 9
1 1 4 4 8 8
We are supposed to insert
an item value 5 and find an
appropriate node z for it

6 6
x=T.root
2 2 9 9
1 1 4 4 8 8
Y=NIL and x=T.root

6 6
x=T.root
2 2 9 9
1 1 4 4 8 8
Now x≠NIL and z.key<x.key
That is [5<6]

6 6
2 2 9 9
x
1 1 4 4 8 8
Y=T.root and x=T.root.left
And z.key>x.key[5>2]
y

6 6
2 2 9 9
x
1 1 4 4 8 8
Y=T.root.left and
x=T.root.left.right
And z.key>x.key[5>4]
y

6 6
2 2 9 9
y
1 1 4 4 8 8
Y=T.root.left.right and x=NIL
Then z.p=T.root.left.right

6 6
2 2 9 9
y
1 1 4 4 8 8
Y=T.root.left.right [y ≠ NIL]
And z.key>y.key [5>4]
So y.right=z that is
T.root.left.right.right=z
z 5

Insertion in BST – Pseudo code
if tree is empty
create a root node with the new key
else
compare key with the top node
if key = node key
replace the node with the new value
else if key > node key
compare key with the right subtree:
if subtree is empty create a leaf node
else add key in right subtree
else key < node key
compare key with the left subtree:
if the subtree is empty create a leaf node
else add key to the left subtree

Insertion into a BST: C code
void insert (ptnode *node, int key)
{
ptnode ptr,
temp = search(*node, key);
if (temp || !(*node)) {
ptr = (ptnode) malloc(sizeof(tnode));
if (IS_FULL(ptr)) {
fprintf(stderr, “The memory is fulln”);
exit(1);
}
ptr->key = key;
ptr->left = ptr->right = NULL;
if (*node)
if (key<temp->key) temp->left=ptr;
else temp->right = ptr;
else *node = ptr;
}
}

Delete a value from the BST
 Removing a node from a BST is a bit more complex, since we do
not want to create any "holes" in the tree. The intention is to
remove the specified item from the BST and adjusts the tree
 The binary search algorithm is used to locate the target item:
starting at the root it probes down the tree till it finds the
target or reaches a leaf node (target not in the tree)
 If the node has one child then the child is spliced to the parent
of the node. If the node has two children then its successor has
no left child; copy the successor into the node and delete the
successor instead TREE-DELETE (T, z) removes the node
pointed to by z from the tree T. IT returns a pointer to the node
removed so that the node can be put on a free-node list
 The overall strategy for deleting a node or item from a binary
search tree can be described through some cases.

Experimenting the cases:
 if the tree is empty return false
 else Attempt to locate the node containing the target
using the binary search algorithm:
if the target is not found return false
else the target is found, then remove its node. Now
while removing the node four cases may happen

Case 1: if the node has 2 empty subtrees
- replace the link in the parent with null
Case 2: if the node has a left and a right subtree
- replace the node's value with the max value in the
left subtree
- delete the max node in the left subtree
Case 3: if the node has no left child
- link the parent of the node to the right
(non-empty) subtree
Case 4: if the node has no right child
- link the parent of the target to the left
(non-empty) subtree

Case 1: removing a node with 2 EMPTY SUBTREES
-replace the link in the parent with null
Parent
Z
Removing 4
5 5
4 4
7 7
9 9
66 88 1100
7 7
5 5
6 6
9 9
88 1100

Case 2: removing a node with 2 SUBTREES
- replace the node's value with the max value in the left subtree
- delete the max node in the left subtree
Z Removing 7
5 5
4 4
7 7
9 9
66 88 1100
6 6
5 5
4 4 8 8
9 9
1100

Case 3: if the node has no left child
- link the parent of the node to the right (non-empty) subtree
Z
Removing 5
5 5
7 7
9 9
66 88 1100
7 7
8 8
9 9
1100
Parent Parent
6 6

Case 4: if the node has no right child
- link the parent of the node to the left (non-empty) subtree
Z
Removing 5
5 5
7 7
9 9
88 1100
7 7
8 8
9 9
1100
Parent Parent
4 4
4 4

BST Deletion Algorithm
• TREE-DELETE (T, z)
1. if left [z] = NIL .OR. right[z] = NIL
2. then y ← z
3. else y ← TREE-SUCCESSOR (z)
4. if left [y] ≠ NIL
5. then x ← left[y]
6. else x ← right [y]
7. if x ≠ NIL
8. then p[x] ← p[y]
9. if p[y] = NIL
10. then root [T] ← x
11. else if y = left [p[y]]
12. then left [p[y]] ← x
13. else right [p[y]] ← x
14. if y ≠ z
15. then key [z] ← key [y]
16. if y has other field, copy them, too
17. return y

Analysis of BST Operations
 The complexity of operations search, insert and
remove in BST is O(h) , where h is the height.
 When the tree is balanced then it is O(log n). The
updating operations cause the tree to become
unbalanced.
 The tree can degenerate to a linear shape and the
operations will become O (n)

Applications for BST
 Binary Search Tree: Used in many search applications where data is
constantly entering/leaving, such as the map and set objects in many
languages' libraries.
 Binary Space Partition: Used in almost every 3D video game to
determine what objects need to be rendered.
 Binary Tries: Used in almost every high-bandwidth router for storing
router-tables.
 Heaps: Used in implementing efficient priority-queues. Also used in
heap-sort.
 Huffman Coding Tree:- used in compression algorithms, such as
those used by the .jpeg and .mp3 file-formats.
 GGM Trees - Used in cryptographic applications to generate a tree of
pseudo-random numbers.
 Syntax Tree: Constructed by compilers and (implicitly) calculators to
parse expressions.

What is a Degenerate BST?
 A degenerate binary search tree is one where most or
all of the nodes contain only one sub node.
 It is unbalanced and, in the worst case, performance
degrades to that of a linked list.
 If your add node function does not handle rebalancing,
then you can easily construct a degenerate tree by
feeding it data that is already sorted.

Does the order of inserting
elements into a tree matter?
 Yes, certain orders might produce very
unbalanced trees or degenerated trees!
 Unbalanced trees are not desirable because search
time increases!
 Advanced tree structures, such as red-black
trees, AVL trees, guarantee balanced trees.

Does the
order of
inserting
elements
into a tree
matter?

Better Search Trees
Prevent the degeneration of the BST :
 A BST can be set up to maintain balance during updating
operations (insertions and removals)
 Types of ST which maintain the optimal performance in
other words balanced trees:
– splay trees
– AVL trees
– 2-4 Trees
– Red-Black trees
– B-trees

References
 Introduction to Algorithms by Thomas H. Cormen
and others
 Binary Search Tree by Penelope Hofsdal
 Rada Mihalcea
http://www.cs.unt.edu/~rada/CSCE3110

MD. Shakhawat Hossain
Student of Computer Science & Engineering Dept.
University of Rajshahi

Binary Search Tree

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